Chapter 1: Introduction to Operational Amplifiers

Example 1.1_a

#Example 1.1_a
#The equivalent circuit of the Motorola op-amp MC 1435 is shown in Figure.No-1.2
#Determine the collector current in each transistor and the dc voltage at the
#output terminal

#Variable declaration
Vcc=6                 #Voltage in volts
Vbe5=0.7              #Voltage in volts
Vee=6                 #Voltage in volts
Vbe3=6.7              #Voltage in volts
Vbe6=0.7              #Voltage in volts
Vbe7=0.7              #Voltage in volts
Rc1=6.7*10**3         #Resistance in ohms
Ic1=0                 #initialization

#Calculation
Vc1=Vcc-Rc1*Ic1
Ve4=Vc1-Vbe5
I4=(Ve4+Vee)/(9.1*10**3+5.5*10**3)
Vb3=5.5*10**3*I4-Vee
Ve3=Vb3-Vbe3
Ie3=(Ve3+Vbe3)/3.3*10**3
Ic1=1.08*10**-3/2.765 #Since Ie3=2*Ic1
Vc1=Vcc-Rc1*Ic1
Ve4=Vc1-Vbe5
Ie4=(Ve4+Vee)/(29.2*10**3)
Ic5=Ie4
Vc5=Vcc-3.8*10**3*Ic5
Ve6=Vc5-Vbe6
Ie6=(Ve6+Vee)/(15*10**3)
Ve7=Ve6+Vbe7
I1=(Vcc-Ve7)/400
Ie8=I1
Ve8=-Vee+2*10**3*Ie8

#Result
print "Collector current Ic1 is",round(Ic1*10**3,2),"mA"
print "Voltage Vc1 is",round(Vc1,2),"V" 
print "Voltage Ve4 is",round(Ve4,2),"V"
print "Current Ie4 is",round(Ie4*10**3,3),"mA"
print "Current Ic5 is",round(Ic5*10**3,3),"mA"
print "Voltage Vc5 is",round(Vc5,2),"V"
print "Voltage Ve6 is",round(Ve6,2),"V"
print "Current Ie6 is",round(Ie6*10**3,3),"mA"
print "Voltage Ve7 is",round(Ve7,2),"V"
print "Current I1 is",round(I1*10**3,2),"mA"
print "Current Ie8 is",round(Ie8*10**3,2),"mA"
print "Voltage Ve8 at the output terminal is",round(Ve8,2),"V"

Collector current Ic1 is 0.39 mA
Voltage Vc1 is 3.38 V
Voltage Ve4 is 2.68 V
Current Ie4 is 0.297 mA
Current Ic5 is 0.297 mA
Voltage Vc5 is 4.87 V
Voltage Ve6 is 4.17 V
Current Ie6 is 0.678 mA
Voltage Ve7 is 4.87 V
Current I1 is 2.82 mA
Current Ie8 is 2.82 mA
Voltage Ve8 at the output terminal is -0.35 V

Example 1.1_b

#Example 1.1_b, Figure.No-1.2
#Calculate the Voltage gain of the opamp

#Variable decclaration
Ie1=0.39*10**-3    #Current in amps
Ie4=0.298*10**-3   #Current in amps
Ie6=0.678*10**-3   #Current in amps
Rc1=6.7*10**3      #Resistance in ohms
Rc5=3.8*10**3      #Resistance in ohms
beta_ac=150

#Calculation
re1=(25*10**-3)/Ie1
re2=re1
re4=(25*10**-3)/Ie4
re5=re4
re6=(25*10**-3)/Ie6
k=(Rc1*2*beta_ac*re4)/(Rc1+2*beta_ac*re4)
Ad1=k/re1
k1=(Rc5*beta_ac*(re6+15*10**3))/(Rc5+beta_ac*(re6+15*10**3))
Ad2=k1/(2*re5)
Ad=Ad1*Ad2

#Result
print "Voltage gain of the dual-input,balanced output-differential amplifier is",round(Ad1,2)
print "Voltage gain of the dual-input,unbalanced output-differential amplifier is",round(Ad2,1)
print "Overall gain of the op-amp is",round(Ad,2)

Voltage gain of the dual-input,balanced output-differential amplifier is 82.55
Voltage gain of the dual-input,unbalanced output-differential amplifier is 22.6
Overall gain of the op-amp is 1866.34

Example 1.1_c

#Example 1.1_c, Figure.No-1.2
#Determine the Input resistance of the opamp

#Variable declaration
beta_ac=150
re1=64.1         #Resistance in ohms

#calculation
Ri=2*beta_ac*re1

#result
print "Input resistance Ri is",round(Ri/10**3,2),"kilo ohms"

Input resistance Ri is 19.23 kilo ohms